(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Please see the attachment

2. Relevant equations

Please see the attachment

3. The attempt at a solution

i.)

Normalisation implies that:

[tex] \int_{-\infty}^{\infty} \Psi (x,t)* \Psi (x,t) dx = 1[/tex]

[tex] \int_{-\infty}^{\infty} |a_{0}| ^{2} | \psi _{0} (x) | ^ {2} + |a_{1}| ^{2} | \psi _{1} (x) | ^ {2} dx = 1 [/tex]

After a bit of work I got to this line

[tex] |a_{0}| ^{2} \int_{-\infty}^{\infty} | \psi _{0} (x) | ^ {2} dx + |a_{1}| ^{2} \int_{-\infty}^{\infty} | \psi _{1} (x) | ^ {2} dx = 1 [/tex]

[tex] |a_{0}| ^{2} + |a_{1}| ^{2} = 1 [/tex]

i.e. I found this to be the normalisation condition.

ii.)

First to see if the energy is an eigenvalue

[tex] [E] \Psi (x,t) = i \hbar \dfrac{d}{dt} \left( a _{0} \psi (x) e^{ - \dfrac{i E_{0} t }{ \hbar } } + a _{1} \psi (x) e^{ - \dfrac{i E_{1} t }{ \hbar } } \right) [/tex]

[tex] [E] \Psi (x,t) = E_{0} a _{0} \psi (x) e^{ - \dfrac{i E_{0} t }{ \hbar } } + E_{1} a _{1} \psi (x) e^{ - \dfrac{i E_{1} t }{ \hbar } } [/tex]

So the energy is not an eigenvalue and hence the energy is not a sharp observable.

[tex] \langle E \rangle = \int_{-\infty}^{\infty} \left( a _{0} * \psi (x) * e^{ \dfrac{i E_{0} t }{ \hbar } } + a _{1} * \psi (x) * e^{ \dfrac{i E_{1} t }{ \hbar } } \right) \left( E_{0} a _{0} \psi (x) e^{ - \dfrac{i E_{0} t }{ \hbar } } + E_{1} a _{1} \psi (x) e^{ - \dfrac{i E_{1} t }{ \hbar } } \right) [/tex]

[tex] \langle E \rangle = E_{0} |a_{0}| ^{2} + E_{1} |a_{1}| ^{2} [/tex]

[tex] \langle E \rangle = E_{0} |a_{0}| ^{2} + E_{1} (1 - |a_{0}| ^{2} ) [/tex]

[tex] \langle E \rangle = |a_{0}| ^{2} ( E_{0} - E_{1} ) + E_{1} [/tex]

[tex] \langle E \rangle = |a_{0}| ^{2} ( \dfrac{1}{2} \hbar \omega - \dfrac{3}{2} \hbar \omega ) + \dfrac{3}{2} \hbar \omega [/tex]

[tex] \langle E \rangle = \hbar \omega \left( \dfrac{3}{2} - |a_{0}| ^{2} \right) [/tex]

I'm not 100% sure if this is right.

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# Homework Help: Quantum mechanics problem!

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